1. Definition and direct answer
The Von Mises criterion, also called the distortion-energy or J₂ criterion, converts a stress tensor into one positive equivalent stress. It is mainly used to assess the onset of yielding in ductile materials.
The value is independent of coordinate orientation, but all components entered in one formula must be consistent and evaluated at the same point.
2. General 3D formula in Cartesian components
σᵥ = √{[(σx−σy)²+(σy−σz)²+(σz−σx)²]/2 + 3(τxy²+τyz²+τzx²)}This expression uses the three normal stresses σx, σy, σz and the three independent shear stresses τxy, τyz, τzx. Every quantity must use the same unit, such as MPa.
Normal-stress signs matter in their differences. A shear-stress sign does not directly change the result because each shear term is squared.
3. Plane-stress 2D formula
σᵥ = √(σx² − σx·σy + σy² + 3τxy²)For a thin plate loaded in its plane, σz = τyz = τzx = 0 is generally assumed. The general formula then reduces to this 2D expression.
One normal stress combined with shear is the special case σy = 0: σᵥ = √(σx² + 3τxy²).
4. Formula from principal stresses
σᵥ = √{[(σ₁−σ₂)²+(σ₂−σ₃)²+(σ₃−σ₁)²]/2}In the principal coordinate system, shear stresses are zero. This form is convenient when σ₁, σ₂ and σ₃ come from an analytical calculation or finite-element software.
Pure hydrostatic stress, with σ₁ = σ₂ = σ₃, gives σᵥ = 0 because it produces no distortion.
5. Axial–bending–torsion case
σx = σa + σb ; σᵥ = √(σx² + 3τ²)At the same section point, first add axial and bending stresses with their signs. Torsion then provides shear stress τ.
Check opposite fibres when bending changes sign: the critical fibre is not always the one chosen intuitively.
6. Quick sanity checks
Three simple limiting cases help detect a formula, sign or unit error.
- If all shear stresses are zero and only one normal stress remains, σᵥ = |σ|.
- If only one shear stress τ remains, σᵥ = √3·|τ|.
- If σ₁ = σ₂ = σ₃, σᵥ = 0.
- The result must not change when every component is expressed in another coherent unit.
7. Limits of use
Comparison with yield strength assumes, among other things, a ductile material, a correctly evaluated stress state and an appropriate design rule. Brittle materials may require another criterion.
Fatigue, welds, plasticity, temperature, stress concentrations, stability and code requirements must be checked separately when relevant.
Numerical application
Preserved example: axial load, bending and torsion
At the same surface point, σa = 80 MPa, σb = 40 MPa and τ = 30 MPa. The allowable value selected for the example is 250 MPa.
- Combine normal stresses: σx = 80 + 40 = 120 MPa.
- Recognize the 2D case with σy = 0: σᵥ = √(σx² + 3τ²).
- Calculate σᵥ = √(120² + 3×30²) = 130.77 MPa.
- Calculate utilization: 130.77/250 = 52.31%.
Result: The simplified factor is approximately 1.912. It does not automatically cover fatigue or stress concentrations.
Formulas checked against MIT OpenCourseWare and NASA resources on the Von Mises criterion. Open exact source ↗
For checking and further study
Direct references
Each link points to the exact course, standard or publication page used, rather than a generic homepage.
- Lecture 17 — Constitutive relations, Von Mises materialMIT OpenCourseWareLecture covering the Von Mises criterion and its stress formulation.↗
- NASA/CR-2001-211162 — yield criteriaNASA Technical Reports ServerTechnical report presenting the Von Mises J₂ criterion and combined-stress cases.↗
- The International System of Units, 9th editionInternational Bureau of Weights and MeasuresOfficial reference for units and symbols.↗